Answer:
5.758 is the density of the metal ingot in grams per cubic centimeter.
Explanation:
1) Mass of pycnometer = M = 27.60 g
Mass of pycnometer with water ,m= 45.65 g
Density of water at 20 °C = d =
1 kg = 1000 g


Mass of water ,m'= m - M = 45.65 g - 27.60 g =18.05 g
Volume of pycnometer = Volume of water present in it = V


2) Mass of metal , water and pycnometer = 56.83 g
Mass of metal,M' = 9.5 g
Mass of water when metal and water are together ,m''= 56.83 g - M'- M
56.83 g - 9.5 g - 27.60 g = 19.7 g
Volume of water when metal and water are together = v

Density of metal = d'
Volume of metal = v' =
Difference in volume will give volume of metal ingot.
v' = v - V


Since volume cannot be in negative .
Density of the metal =d'
=
Molecular are only between non-metals. Ionic has a higher melting point
Answer:
643g of methane will there be in the room
Explanation:
To solve this question we must, as first, find the volume of methane after 1h = 3600s. With the volume we can find the moles of methane using PV = nRT -<em>Assuming STP-</em>. With the moles and the molar mass of methane (16g/mol) we can find the mass of methane gas after 1 hour as follows:
<em>Volume Methane:</em>
3600s * (0.25L / s) = 900L Methane
<em>Moles methane:</em>
PV = nRT; PV / RT = n
<em>Where P = 1atm at STP, V is volume = 900L; R is gas constant = 0.082atmL/molK; T is absolute temperature = 273.15K at sTP</em>
Replacing:
PV / RT = n
1atm*900L / 0.082atmL/molK*273.15 = n
n = 40.18mol methane
<em>Mass methane:</em>
40.18 moles * (16g/mol) =
<h3>643g of methane will there be in the room</h3>